symmetric monoidal (∞,1)-category of spectra
The determinant is the (essentially unique) universal alternating multilinear map.
Let Vect${}_k$ be the category of vector spaces over a field $k$, and assume for the moment that the characteristic $char(k) \neq 2$. For each $j \geq 0$, let
be the 1-dimensional sign representation on the symmetric group $S_j$, taking each transposition $(i j)$ to $-1 \in k^\ast$. We may linearly extend the sign action of $S_j$, so that $sgn$ names a (right) $k S_j$-module with underlying vector space $k$. At the same time, $S_j$ acts on the $j^{th}$ tensor product of a vector space $V$ by permuting tensor factors, giving a left $k S_j$-module structure on $V^{\otimes j}$. We define the Schur functor
by the formula
It is called the $j^{th}$ alternating power (of $V$).
Another point of view on the alternating power is via superalgebra. For any cosmos $\mathbf{V}$ let $CMon(\mathbf{V})$ be the category of commutative monoid objects in $\mathbf{V}$. The forgetful functor $CMon(\mathbf{V}) \to \mathbf{V}$ has a left adjoint
whose values are naturally regarded as graded by degree $n$.
This applies in particular to $\mathbf{V}$ the category of supervector spaces; if $V$ is a supervector space concentrated in odd degree, say with component $V_{odd}$, then each symmetry $\sigma: V \otimes V \to V \otimes V$ maps $v \otimes w \mapsto -w \otimes v$ for elements $v, w \in V_{odd}$. It follows that the graded component $\exp(V)_n$ is concentrated in $parity(n)$ degree, with component $\Lambda^n(V_{odd})$.
There is a canonical natural isomorphism $\Lambda^n(V \oplus W) \cong \sum_{j + k = n} \Lambda^j(V) \otimes \Lambda^k(W)$.
Again take $\mathbf{V}$ to be the category of supervector spaces. Since the left adjoint $\exp: \mathbf{V} \to CMon(\mathbf{V})$ preserves coproducts and since the tensor product $\otimes$ of $\mathbf{V}$ provides the coproduct for commutative monoid objects, we have a natural isomorphism
Examining the grade $n$ component $\exp(V \oplus W)_n$, this leads to an identification
and now the result follows by considering the case where $V, W$ are concentrated in odd degree.
If $V$ is $n$-dimensional, then $\Lambda^j(V)$ has dimension $\binom{n}{j}$. In particular, $\Lambda^n(V)$ is 1-dimensional.
By induction on dimension. If $\dim(V) = 1$, we have that $\Lambda^0(V)$ and $\Lambda^1(V)$ are $1$-dimensional, and clearly $\Lambda^n(V) = 0$ for $n \geq 2$, at least when $char(k) \neq 2$.
We then infer
where the dimensions satisfy the same recurrence relation as for binomial coefficients: $\binom{n+1}{j} = \binom{n}{j} + \binom{n}{j-1}$.
More concretely: if $e_1, \ldots, e_n$ is a basis for $V$, then expressions of the form $e_{n_1} \otimes \ldots \otimes e_{n_j}$ form a basis for $V^{\otimes j}$. Let $e_{n_1} \wedge \ldots \wedge e_{n_j}$ denote the image of this element under the quotient map $V^{\otimes j} \to \Lambda^j(V)$. We have
(consider the transposition in $S_j$ which swaps $i$ and $i+1$) and so we may take only such expressions on the left where $n_1 \lt \ldots \lt n_j$ as forming a spanning set for $\Lambda^j(V)$, and indeed these form a basis. The number of such expressions is $\binom{n}{j}$.
In the case where $char(k) = 2$, the same development may be carried out by simply decreeing that $e_{n_1} \wedge \ldots \wedge e_{n_j} = 0$ whenever $n_i = n_{i'}$ for some pair of distinct indices $i$, $i'$.
Now let $V$ be an $n$-dimensional space, and let $f \colon V \to V$ be a linear map. By the proposition, the map
being an endomorphism on a 1-dimensional space, is given by multiplying by a scalar $D(f) \in k$. It is manifestly functorial since $\Lambda^n$ is, i.e., $D(f g) = D(f) D(g)$. The quantity $D(f)$ is called the determinant of $f$.
We see then that if $V$ is of dimension $n$,
is a homomorphism of multiplicative monoids; by commutativity of multiplication in $k$, we infer that
for each invertible linear map $U \in GL(V)$.
If we choose a basis of $V$ so that we have an identification $GL(V) \cong Mat_n(k)$, then the determinant gives a function
that takes products of $n \times n$ matrices to products in $k$. The determinant however is of course independent of choice of basis, since any two choices are related by a change-of-basis matrix $U$, where $A$ and its transform $U A U^{-1}$ have the same determinant.
By following the definitions above, we can give an explicit formula:
We work over fields of arbitrary characteristic. The determinant satisfies the following properties, which taken together uniquely characterize the determinant. Write a square matrix $A$ as a row of column vectors $(v_1, \ldots, v_n)$.
$\det$ is separately linear in each column vector:
$\det(v_1, \ldots, v_n) = 0$ whenever $v_i = v_j$ for distinct $i, j$.
$\det(I) = 1$, where $I$ is the identity matrix.
Other properties may be worked out, starting from the explicit formula or otherwise:
If $A$ is a diagonal matrix, then $\det(A)$ is the product of its diagonal entries.
More generally, if $A$ is an upper (or lower) triangular matrix, then $\det(A)$ is the product of the diagonal entries.
If $E/k$ is a field extension and $f$ is a $k$-linear map $V \to V$, then $\det(f) = \det(E \otimes_k f)$. Using the preceding properties and the Jordan normal form? of a matrix, this means that $\det(f)$ is the product of its eigenvalues (counted with multiplicity), as computed in the algebraic closure of $k$.
If $A^t$ is the transpose of $A$, then $\det(A^t) = \det(A)$.
A simple observation which flows from these basic properties is
(Cramer’s Rule)
Let $v_1, \ldots, v_n$ be column vectors of dimension $n$, and suppose
Then for each $i$ we have
where $w$ occurs as the $i^{th}$ column vector on the right.
This follows straightforwardly from properties 1 and 2 above.
For instance, given a square matrix $A$ such that $\det(A) \neq 0$, and writing $A = (v_1, \ldots, v_n)$, this allows us to solve for a vector $a$ in an equation
and we easily conclude that $A$ is invertible if $\det(A) \neq 0$.
This holds true even if we replace the field $k$ by an arbitrary commutative ring $R$, and we replace the condition $\det(A) \neq 0$ by the condition that $\det(A)$ is a unit. (The entire development given above goes through, mutatis mutandis.)
Given a linear endomorphism $f: M\to M$ of a finite rank free unital module over a commutative unital ring, one can consider the zeros of the characteristic polynomial $\det(t \cdot 1_V - f)$. The coefficients of the polynomial are the concern of the Cayley-Hamilton theorem.
A useful intuition to have for determinants of real matrices is that they measure change of volume. That is, an $n \times n$ matrix with real entries will map a standard unit cube in $\mathbb{R}^n$ to a parallelpiped in $\mathbb{R}^n$ (quashed to lie in a hyperplane if the matrix is singular), and the determinant is, up to sign, the volume of the parallelpiped. It is easy to convince oneself of this in the planar case by a simple dissection of a parallelogram, rearranging the dissected pieces in the style of Euclid to form a rectangle. In algebraic terms, the dissection and rearrangement amount to applying shearing or elementary column operations to the matrix which, by the properties discussed earlier, leave the determinant unchanged. These operations transform the matrix into a diagonal matrix whose determinant is the area of the corresponding rectangle. This procedure easily generalizes to $n$ dimensions.
The sign itself is a matter of interest. An invertible transformation $f \colon V \to V$ is said to be orientation-preserving if $\det(f)$ is positive, and orientation-reversing if $\det(f)$ is negative. Orientations play an important role throughout geometry and algebraic topology, for example in the study of orientable manifolds (where the tangent bundle as $GL(n)$-bundle can be lifted to a $GL_+(n)$-bundle structure, $GL_+(n) \hookrightarrow GL(n)$ being the subgroup of matrices of positive determinant). See also KO-theory.
Finally, we include one more property of determinants which pertains to matrices with real coefficients (which works slightly more generally for matrices with coefficients in a local field):
If $A$ is an $n \times n$ matrix, the determinant of its exponential equals the exponential of its trace
More generally, the determinant of $A$ is a polynomial in the traces of the powers of $A$:
For $2 \times 2$-matrices:
For $3 \times 3$-matrices:
For $4 \times 4$-matrices:
Generally for $n \times n$-matrices (Kondratyuk-Krivoruchenko 92, appendix B):
It is enough to prove this for semisimple matrices $A$ (matrices that are diagonalizable upon passing to the algebraic closure of the ground field) because this subset of matrices is Zariski dense (using for example the nonvanishing of the discriminant of the characteristic polynomial) and the set of $A$ for which the equation holds is Zariski closed.
Thus, without loss of generality we may suppose that $A$ is diagonal with $n$ eigenvalues $\lambda_1, \ldots, \lambda_n$ along the diagonal, where the statement can be rewritten as follows. Letting $p_k = tr(A^k) = \lambda_1^k + \ldots + \lambda_n^k$, the following identity holds:
This of course is just a polynomial identity, one closely related to various of the Newton identities? that concern symmetric polynomials in indeterminates $x_1, \ldots, x_n$. Thus we again let $p_k = x_1^k + \ldots + x_n^k$, and define the elementary symmetric polynomials $\sigma_k = \sigma_k(x_1, \ldots, x_n)$ via the generating function identity
Then we compute
and simply match coefficients of $t^n$ in the initial and final series expansions, where we easily compute
This completes the proof.
see Pfaffian for the moment
matrix, linear algebra, exterior algebra, characteristic polynomial
quasideterminant, Berezinian,Jacobian, Pfaffian, hafnian, Wronskian
determinant line, determinant line bundle, Pfaffian line bundle
See also
One derivation of the formula (1) for the determinant as a polynomial in traces of powers is spelled out in appendix B of
Last revised on October 21, 2018 at 09:41:41. See the history of this page for a list of all contributions to it.